Let $\varphi$be the standard normal distribution function, and let X be a normal random variable with mean μ and variance 1. We want to find E[ $\varphi$(X)]. To do so, let Z be a standard normal random variable that is independent of X, and let

$I=\left\{\begin{array}{l}1,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if}Z<X\\ 0,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if}Z\ge X\end{array}\right.$

(a) Show that $E[I\mid X=x]=\Phi \left(x\right)$.

(b) Show that $E\left[\Phi \right(X\left)\right]=P\{Z<X\}$.

(c) Show that $E\left[\Phi \right(X\left)\right]=\Phi \left(\frac{\mu }{\sqrt{2}}\right)$.

Hint: What is the distribution of $X-Z$?

The preceding comes up in statistics. Suppose you are about to observe the value of a random variable X that is normally distributed with an unknown mean μ and variance 1, and suppose that you want to test the hypothesis that the mean μ is greater than or equal to 0. Clearly you would want to reject this hypothesis if X is sufficiently small. If it results that X = x, then the p-value of the hypothesis that the mean is greater than or equal to 0 is defined to be the probability that X would be as small as x if μ were equal to 0 (its smallest possible value if the hypothesis were true). (A small p-value is taken as an indication that the hypothesis is probably false.) Because X has a standard normal distribution when μ = 0, the p-value that results when X = x is $\varphi$ (x). Therefore, the preceding shows that the expected p-value that results when the true mean is μ is $\phi \left(\frac{\mu }{\sqrt{2}}\right)$ .

Step by step solution

01

Given information (Part a)

$\varphi =$standard normal distribution function.

$X=$normal random variable with mean $\mu$and variance 2

role="math" localid="1647418706310" $Z=$standard normal random variable that is independent of X

$I=\left\{\begin{array}{c}1,\text{if}\mathrm{Z}<\mathrm{X}\\ 0,\text{if}Z\ge \mathrm{X}\end{array}\right.$

02

Solution (Part a)

We are going to show that $E[I\mid X=x]=\mathrm{\Phi }\left(x\right)$

So,

$E[I\mid X=x]=(1\times P\{Z<X\mid X=x\left\}\right)+(0\times P\{Z\ge X\mid X=x\left\}\right)$

$=P\{Z<x\mid X=x\}$

$=P\{Z<x\}$

Now take the given definition as $\Phi \left(x\right)$is standard normal distribution function, one has that:

So that,

$P\{Z<x\}=\mathrm{\Phi }\left(x\right)$

$E[I\mid X=x]=\mathrm{\Phi }\left(x\right)$

03

Final answer (Part a)

It has been now shown that the$E[I\mid X=x]=\mathrm{\Phi }\left(x\right)$.

04

Given information (Part b)

$\varphi =$standard normal distribution function.

$X=$normal random variable with mean $\mu$and variance 2

$Z=$standard normal random variable that is independent of X

$I=\left\{\begin{array}{c}1,\text{if}\mathrm{Z}<\mathrm{X}\\ 0,\text{if}\mathrm{Z}\ge \mathrm{X}\end{array}\right.$

05

Solution (Part b)

Here we need to show that $E\left[\mathrm{\Phi }\right(x\left)\right]=P\{Z<X\}$

From the above result,

$E[I\mid X]=\mathrm{\Phi }\left(X\right)$

Moreover, one has the following relations:

$E\left[I\right]=E\left[E\right[I\mid X\left]\right]$

$=E\left[\mathrm{\Phi }\right(X\left)\right]-----\left(1\right)$

So,

$E\left[I\right]=\left(P\right\{I=1\}\times 1)+(0\times P\{I=0\left\}\right)$

$=P\{Z<X\}\times 1+P\{Z\ge X\}\times 0$

$=P\{Z<X\}------\left(2\right)$

Now we need to combine The equation (1) and (2) That will get.

$E\left[\mathrm{\Phi }\right(X\left)\right]=P\{Z<X\}$

06

Final answer (Part b)

Now it has been shown that$E\left[\mathrm{\Phi }\right(X\left)\right]=P\{Z<X\}$.

07

Given information (Part c)

$\varphi =$standard normal distribution function.

$X=$normal random variable with mean $\mu$and variance 2

$Z=$standard normal random variable that is independent of X

$I=\left\{\begin{array}{c}1,\text{if}Z<\mathrm{X}\\ 0,\text{if}Z\ge \mathrm{X}\end{array}\right.$

08

Solution (Part c)

We need to show that $E\left[\mathrm{\Phi }\right(x\left)\right]=\mathrm{\Phi }\left(\frac{\mu }{\sqrt{2}}\right)$

If, X-Z is normal with mean $\mu$and variance 2 .

$P\{X>Z\}=P\{X-Z>0\}$

$=P\left\{\frac{X-Z-\mu }{\sqrt{2}}>\frac{-\mu }{\sqrt{2}}\right\}$

$=1-\mathrm{\Phi }\left(\frac{-\mu }{\sqrt{2}}\right)$

$=\mathrm{\Phi }\left(\frac{\mu }{\sqrt{2}}\right)$

Therefore the result is,

$E\left[\mathrm{\Phi }\right(X\left)\right]=\mathrm{\Phi }\left(\frac{\mu }{\sqrt{2}}\right)$

09

Final answer (Part c)

Now it has been shown that$E\left[\mathrm{\Phi }\right(X\left)\right]=\mathrm{\Phi }\left(\frac{\mu }{\sqrt{2}}\right)$.

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